An equilateral triangle and a square are inscribed in a circle as shown. $ABC$ is isosceles. The triangle and square share a common vertex. What is the number of degrees in the measure of the angle indicated by the question mark?

[asy]

import markers; defaultpen(linewidth(0.8));

pair A,B,C,D,E,F,G;

draw(unitcircle);

A=(-1,0);
B=(0,-1);
C=(1,0);
D=(0,1);

draw(A--B--C--D--A);

E=(-0.9,-0.45);
F=(0.9,-0.45);

draw(D--E--F--D);

G=(-0.76,-0.23);

markangle(Label("?"),n=1,radius=10,D,G,A,marker(stickframe(n=0),true));

label("$A$",(-0.5,-0.34));
label("$B$",(0.5,-0.34));
label("$C$",B,S);

[/asy]
We can determine the other two angles in the triangle with the unknown angle.  Label its vertices $X$, $Y$, and $Z$.

[asy]
import markers; defaultpen(linewidth(0.8));

pair A,B,C,D,EE,F,G;

draw(unitcircle);

A=(-1,0);
B=(0,-1);
C=(1,0);
D=(0,1);

draw(A--B--C--D--A);

EE=(-0.9,-0.45);
F=(0.9,-0.45);

draw(D--EE--F--D);

G=(-0.76,-0.23);

markangle(Label("?"),n=1,radius=10,D,G,A,marker(stickframe(n=0),true));

draw(A--D--G--A,red+1bp);

label("$X$",D,N);
label("$Y$",A,W);
label("$Z$",G,E);

[/asy]

We want to find $\angle XZY$.  Since $\angle XYZ$ is an angle of a square, $\angle XYZ=90^\circ$.  Also, $\angle YXZ$ is part of an angle of a square.  By symmetry, the large angle at $X$ can be dissected into a sum of three angles, \[90^\circ = \angle YXZ +60^\circ+\angle YXZ=2\angle YXZ+60^\circ.\] Therefore $\angle YXZ=15^\circ$.  The mystery angle is the third angle of this triangle, so \[\angle XZY=180^\circ-90^\circ-15^\circ=\boxed{75^\circ}.\]